Puzzle: Crossing the English Channel

TheOriginalHappyGoat

Moderator
Moderator
Oct 4, 2010
57,360
30,536
113
Margaritaville
Haven't posted one of these in a while, but Alex Bellos had a fun little puzzle today that wasn't too mentally taxing:

Six friends – Babs, Charles, Hattie, Joan, Kenneth and Sid – are going camping in France.

They are travelling across the Channel on a magic carpet (of course) that can take only two people at a time. So, in order for everyone to get across there will need to be 9 trips in total from their starting point in England: 5 across carrying two people, and 4 returning carrying one person. (Magic carpets cannot fly with no one on board, which is why one person needs to travel back. All the friends are able to fly alone on the carpet if need be.)

The faster the carpet flies, the bumpier the ride. Babs can fly at a speed that gets her to France in 1 minute. Charles feels queasy at that speed: his maximum speed gets him to France in 5 minutes. The others can get to France without suffering debilitating travel sickness no faster than 6 mins (Hattie), 7 mins (Joan), 8 mins (Kenneth) and 12 mins (Sid).

1. What is the least amount of flying time, in minutes, required to get everyone from England to France without anyone suffering from queasiness? (When two people are on the carpet together, the crossing will take as long as it takes the slowest of the two.)

2. Let’s say that the friends’ tolerance for travel-sickness means that their fastest journeys take the following times: Babs (1 minute), Charles (3 mins), Hattie (7 mins), Joan (9 mins), Kenneth (11 mins) and Sid (13 mins).

Now what is the least time, in minutes, required to fly all friends to France?

Here’s a tip: the answer to question 1 is less than 45 minutes. And the answer to question 2 is less than the answer to 1.​

As per usual, answers in spoilers, please.
 

Marvin the Martian

Hall of Famer
Sep 4, 2001
29,453
13,056
113
I believe the secret is to get the two slowest over once at the same time. I used the 1st initial for each

B 1
C 5
H 6
J 7
K 8
S 12

B and C over 5
B back 1
S and K over 12
C back 5
H and B over 6
B back 1
J and B over 7
B back 1
B and C over 5

Total time 43
 

sdhoosier

All-Big Ten
Dec 21, 2001
3,804
6,551
113
So Calif
2v1ifg.gif
 

76-1

All-American
Mar 22, 2017
7,474
10,867
113
Indiana
Unless they've hacked the French Air and Beach Defense systems they'll be downed by a Roland AA missile (SAM) within seconds of entering the French Air I D zone...

So the answer is: at least two individuals and one carpet will be destroyed in a flaming spiral and no one completes their flight...;)
 

Marvin the Martian

Hall of Famer
Sep 4, 2001
29,453
13,056
113
Unless they've hacked the French Air and Beach Defense systems they'll be downed by a Roland AA missile (SAM) within seconds of entering the French Air I D zone...

So the answer is: at least two individuals and one carpet will be destroyed in a flaming spiral and no one completes their flight...;)

Or the more likely possibility, France surrenders to the 2 people on a carpet.
 

TheOriginalHappyGoat

Moderator
Moderator
Oct 4, 2010
57,360
30,536
113
Margaritaville
I believe the secret is to get the two slowest over once at the same time. I used the 1st initial for each

B 1
C 5
H 6
J 7
K 8
S 12

B and C over 5
B back 1
S and K over 12
C back 5
H and B over 6
B back 1
J and B over 7
B back 1
B and C over 5

Total time 43
Nope.

You have the right strategy for Q2, but not Q1.
 

outside shooter

Hall of Famer
Gold Member
Oct 23, 2001
20,367
7,790
113
B 1
C 5
H 6
J 7
K 8
S 12

B and C over 5
B back 1
B and H over 6
B back 1
B and J over 7
B back 1
B and K over 8
B back 1
B and S over 12

Total time 42
 

Noodle

Hall of Famer
Jun 19, 2001
26,062
5,552
113
First one:
B and S over 12
B back 1
B and K over 8
B back 1
B and J over 7
B back 1
B and H over 6
B back 1
B and C over 5
Total = 42 minutes (doesn't matter the order, as long as B goes on every trip)

Second one:
B and C over 3
C back 3
K and S over 13
B back 1
B and J over 9
B back 1
B and H over 7
B back 1
B and C over 3
Total = 41 minutes (you could do this in a variety of other ways as long as K and S make a trip together after B and C go over, with C making two trips over and one back, and B making four trips over and three back)
 

TheOriginalHappyGoat

Moderator
Moderator
Oct 4, 2010
57,360
30,536
113
Margaritaville
First one:
B and S over 12
B back 1
B and K over 8
B back 1
B and J over 7
B back 1
B and H over 6
B back 1
B and C over 5
Total = 42 minutes (doesn't matter the order, as long as B goes on every trip)

Second one:
B and C over 3
C back 3
K and S over 13
B back 1
B and J over 9
B back 1
B and H over 7
B back 1
B and C over 3
Total = 41 minutes (you could do this in a variety of other ways as long as K and S make a trip together after B and C go over, with C making two trips over and one back, and B making four trips over and three back)
Good but you can actually make the second one even shorter.